poj 1061 青蛙的约会(扩展gcd)

题目链接

题意：两只青蛙从数轴正方向跑，给出各自所在位置， 和数轴长度，和各自一次跳跃的步数，问最少多少步能相遇。

分析：(x+m*t) - (y+n*t) = p * L;(t是跳的次数，L是a青蛙跳的圈数跟b青蛙的圈数之差。整个就是路程差等于纬度线周长的整数倍)。

（x+m*t）- (y+n*t) = p*L;

(n-m)*t  + p*L = x - y;

令a = n-m; b = L; c = x-y;  d = gcd(a, b);

a *t  + b*p = c;

这道题的思路都是看的这个博客 ， 这个博客里有这个题目的推导，还有扩展gcd的推导。

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#define LL long long
using namespace std;

void exgcd(LL a, LL b, LL &d, LL &x, LL &y)
{
    if(!b) {d = a; x = 1; y = 0;}
    else{ exgcd(b, a%b, d, y, x); y -= x*(a/b); }
}
int main()
{
    LL a, b, d, c;
    LL x, y, m, n, l;
    cin>>x>>y>>m>>n>>l;
    a = n-m;
    b = l;
    c = x - y;
    exgcd(a, b, d, x, y);
    if(c%d!=0)
    printf("Impossible
");
    else
    {
        LL q;
        q = b/d;
        printf("%lld
", (x*(c/d)%q+q)%q);
    }

    return 0;
}

扩展gcd模板：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#define LL long long
using namespace std;

void exgcd(LL a, LL b, LL &d, LL &x, LL &y)
{
    if(!b) {d = a; x = 1; y = 0;}
    else{ exgcd(b, a%b, d, y, x); y -= x*(a/b); } //一定注意括号（a/b）;
}
int main()
{
    LL a, b, d, x, y;
    while(cin>>a>>b)
    {
        exgcd(a, b, d, x, y);
        printf("%lld %lld %lld %lld %lld
", a, x, b, y, d);
    }
    return 0;
}